Lorenz Gauge Condition in Helmholtz equation

Question

I do not understand why we can apply the Lorenz Gauge Condition in Helmholtz equation. What is its physical meaning? Any help is appreciated.

$\nabla\cdot \vec{A} + \mu_0\varepsilon_0\frac{∂\phi}{∂t} = 0$

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SOLUTION:

We can describe electromagnetic fields as a function of the electromagnetic potentials through the following equations (delayed potential equations):

$\vec{B}=\nabla × \vec{A}$

$\vec{E}=-\nabla \phi - \frac{∂\vec{A}}{∂t}$

We can transform these electromagnetic potentials so that they continue fulfilling the previous equations. Making the following changes we will obtain the same electromagnetic fields but the potentials will lose their physical meaning. These transformations will allow us to solve simpler statements, like Helmholtz's equation.

Then we can rewrite the new potentials like:

$\vec{A'}=\vec{A}+\vec{\alpha}$

$\phi'=\phi+\beta$

As I mentioned before, these transformations will have to satisfy the delayed potential equations so:

$\vec{B}=\nabla × \vec{A'}=\nabla × (\vec{A}+\vec{\alpha})$

$\vec{E}=-\nabla \phi' - \frac{∂\vec{A'}}{∂t}=-\nabla (\phi+\beta)- \frac{∂(\vec{A}+\vec{\alpha})}{∂t}$

If we continue simplifying the following equations we obtain:

$\nabla ×\vec{\alpha}=0$; $\vec{\alpha}=\nabla\lambda$; $\beta=\frac{∂\lambda}{∂t}$

So finally we get:

$\vec{A'}=\vec{A}+\nabla\lambda$

$\phi'=\phi+\frac{∂\lambda}{∂t}$

Conclusion: we can find any electromagnetic potential that satisfies Maxwell's equations. So we will try to find potentials that ease the calculations. It is important to note that these transformations will lead to a loss of potencial's physical meaning. But we do not have to worry about the electromagnetic fields because they will remain the same.

Answer 1

The Maxwell equations are gauge invariant. If you choose the Lorenz gauge, instead of the free Maxwell equations you get an equation saying that the d'Alembertian of the potential vanishes. If then you choose a complex solution with temporal dependence $\exp(j\omega t)$ (I am not sure about the sign of the phase though), you get the Helmholtz equation.

Answer 2

Assuming $A$ is the vector potential and $\phi=\partial_{t}V$ is the scalar potential, the potentials are calculation aids. The physical fields are the $E$ and $B$ fields. Choosing the Lorenz gauge allows one to solve the coupled partial $2$-nd order differential equation for $A$ and $V$ by uncoupling $A$ and $V$ into two independent partial $2$-nd order differential equations.